Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON):

Solution:
import sys

def reconstruct(rows, cols):
    matrix = [[x for x in range(len(cols))] for y in range(len(rows))]
    for rid in range(len(rows)):
        for cid in range(len(cols)):
            matrix[rid][cid] = rows[rid] + cols[cid]
    return matrix

def reconcile(matrix, rows, cols):
    f_rows = [row for row in rows]
    s_cols = [col for col in cols]
#    f_matrix = [[matrix[x][y] for y in range(len(cols))] for x in range(len(rows))]
#    s_matrix = [[matrix[x][y] for y in range(len(cols))] for x in range(len(rows))]


    for row in range(len(rows)):
        val = -1
        for col in range(len(cols)):
            if val == -1:
                val = matrix[row][col]
            elif val != matrix[row][col]:
                val = -1
                break
        if val != -1:
            f_rows[row] -= val

    for col in range(len(cols)):
        val = -1
        for row in range(len(rows)):
            if val == -1:
                val = matrix[row][col]
            elif val != matrix[row][col]:
                val = -1
                break
        if val != -1:
            s_cols[col] -= val
    if sum(rows) - sum(f_rows) > sum(cols) - sum(s_cols):
        return f_rows, cols
    else:
        return rows, s_cols

def subtract(matrix1, matrix2):
    rows = len(matrix1)
    cols = len(matrix1[0])
    matrix = [[x for x in range(cols)] for y in range(rows)]
    for x in range(rows):
        for y in range(cols):
            matrix[x][y] = matrix1[x][y] - matrix2[x][y]
    return matrix

def pmatrix(matrix):
    for row in matrix:
        print row

def solve(grid):
    rows = [-1 for x in range(len(grid))]
    cols = [-1 for x in range(len(grid[0]))]
    for rid in range(len(grid)):
        for cid in range(len(grid[0])):
            if rows[rid] == -1 or grid[rid][cid] < rows[rid]:
                rows[rid] = grid[rid][cid]
            if cols[cid] == -1 or grid[rid][cid] < cols[cid]:
                cols[cid] = grid[rid][cid]
    matrix = reconstruct(rows, cols)
    sub_matrix = subtract(matrix, grid)
    n_rows, n_cols = reconcile(sub_matrix, rows, cols)
    if reconstruct(n_rows, n_cols) != grid:
        return -1, [], []
    else:
        return sum(n_rows) + sum(n_cols), n_rows, n_cols
    #print "recons"
    #pmatrix(matrix)
    #print "sub"
    #pmatrix(sub_matrix)
    #print "grid"
    #pmatrix(grid)

if __name__ == "__main__":
    n, m = map(int, sys.stdin.readline().split())
    grid = []
    for row in range(n):
        grid.append(map(int, sys.stdin.readline().split()))
    n, r, c = solve(grid)
    print n
    if n != -1:
        for rid, ele in enumerate(r):
            for x in range(ele):
                print "row {}".format(rid + 1)
        for cid, ele in enumerate(c):
            for x in range(ele):
                print "col {}".format(cid + 1)